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Best LeetCode Lists for Interviews

Here is a list of some of the best questions asked in interviews:-  https://github.com/nazarmubeen/TopProblems/blob/master/README.md Must do 75 https://leetcode.com/list/5hkn6wze/ Must do 60  https://leetcode.com/list/5eie1aqd/ Must do medium:-  https://leetcode.com/list/5xaelz7g/ Must do Easy:-   https://leetcode.com/list/5r7rxpr1/ Graph:-  https://leetcode.com/list/x18ervrd/  Dynamic Programming:-    https://leetcode.com/list/x14z0dxr/  FaceBook interviews:- https://leetcode.com/list/xyu98pv6/  Amazon Interviews:-  https://leetcode.com/list/5hkniyf7/  Google Interviews:- https://leetcode.com/list/xyu9xfo1/

DECODE WAYS - LEETCODE PROBLEM 91

  A message containing letters from   A-Z   is being encoded to numbers using the following mapping: 'A' -> 1 'B' -> 2 ... 'Z' -> 26 Given a  non-empty  string containing only digits, determine the total number of ways to decode it. Example 1: Input: "12" Output: 2 Explanation:  It could be decoded as "AB" (1 2) or "L" (12). Example 2: Input: "226" Output: 3 Explanation:  It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6). class Solution { public int numDecodings(String s) { int ways=0; ways=countDecoding(s.toCharArray(),s.length()); return ways; } int countDecoding(char[] digits, int n){ // System.out.println(" digits "+disgits[0]); // for base condition "01123" should return 0 // base cases if (n == 0 || (n == 1 && digits[0]!='

EDIT DISTANCE PROBLEM - LEETCODE 72

  Given two words  word1  and  word2 , find the minimum number of operations required to convert  word1  to  word2 . You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a character Example 1: Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e') Solution:- class Solution { public int minDistance(String word1, String word2) { if(word1.length()==0){ return word2.length(); } if(word2.length()==0){ return word1.length(); } if(word1.length()==0 && word2.length()==0){ return 0; } int[][] result=new int[word1.length()+1][word2.length()+1]; for(int i=0;i<=word1.length();i++){ result[i][0]=i; }

TWO SUM -LEET CODE PROBLEM 1

   https://leetcode.com/problems/two-sum/ Given an array of integers  nums  and an integer  target , return  indices of the two numbers such that they add up to  target . You may assume that each input would have  exactly  one solution , and you may not use the  same  element twice. You can return the answer in any order.   Example 1: Input: nums = [2,7,11,15], target = 9 Output: [0,1] Output: Because nums[0] + nums[1] == 9, we return [0, 1]. Example 2: Input: nums = [3,2,4], target = 6 Output: [1,2] Example 3: Input: nums = [3,3], target = 6 Output: [0,1] class Solution {     public int[] twoSum(int[] nums, int target) {              HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();                  for(int i=0;i<nums.length;i++){                          if(map.get(nums[i])==null){                 map.put(target-nums[i],i);             }             else{                 return new int[]{map.get(nums[i]),i};             }         }         return null

Calculating size of User and Cache storage

user storage:- it can be solved using same scenario as mentioned in first article . https://tech.nazarmubeenworks.com/2019/08/basics-of-system-design-chapter-0.html Suppose there are 12 Million of users are adding every year meaning 1 Million per month. So if we consider 5 year there will be around 60 million of users. Now in terms of data we can look to our DB table and get information about it. A user will be generally having name , id , address , some forien keys . Lets assume 10 columns with each column on an average storing 4 byte of data. 10*4 = 40 bytes of data for one user. 60 * 10^6 * 40 = 2400 * 10^6 = 2.4 * 10^9 = 2.4 GB of data we need to store only user values. Also while calculating storage for the user there is also one important point we need to remember is of ids. If we are going to have 60 Million of users which means 60*10^6 users so unique id’s will be as we know below figures are almost equal

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